The stopping distance for a boat in calm water is modelled by the function d(v) = 0.004v2 + 0.2v + 6, where d(v) is in metres and v is in kilometres per hour.
a. what is the stopping distance if the speed is 10km/h?
b. what is the initial speed of the boat if it takes 11.6m to stop?
: (

a. 8.4 km  b. 20 km/hr or 20,000 m/hr

Step-by-step explanation:

This is your polynomial:

The important thing to realize is that d(v) is the distance it takes for the boat to stop.  That will come later, in part b. Besides that, we also need to remember that v is velocity, which is speed, in km/hr.

For part a. we are looking for d(v), the stopping distance, when v = 10.  That means that we will sub in a 10 for each v in the function and solve for d(v):

so

d(10) = 8.4 km

Now comes the part I was referring to above.  Part b is asking us the speed of the boat if it takes 11.6 meters to stop.  If d(v) is the stopping distance, we sub 11.6 in for d(v) in the function:

The only way w can solve this for velocity is to get everything on one side of the equals sign, set the polynomial equal to 0, then plug the values into the quadratic formula.  

Plugging that into the quadratic formula gives you 2 values of velocity:

v = 20 km/hr and -70 km/hr

We all know that neither time nor distance in math will EVER be negative so we can discount the negative number.  However, I believe that you asked for the distance in meters, so 20 km/hr is the same as 20,000 m/hr.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐